Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 35}{x + 8} = \dfrac{-17x - 37}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 35}{x + 8} (x + 8) = \dfrac{-17x - 37}{x + 8} (x + 8)$ $ x^2 + 35 = -17x - 37$ Subtract $-17x - 37$ from both sides: $ x^2 + 35 - (-17x - 37) = -17x - 37 - (-17x - 37)$ $ x^2 + 35 + 17x + 37 = 0$ $ x^2 + 72 + 17x = 0$ Factor the expression: $ (x + 8)(x + 9) = 0$ Therefore $x = -8$ or $x = -9$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.